On Wednesday 29 August 2007 00:54:19 David Miller wrote:
> From: Michael Buesch <[EMAIL PROTECTED]>
> Date: Tue, 28 Aug 2007 16:48:44 +0200
>
> > On Monday 27 August 2007 23:11:50 David Miller wrote:
> > > From: Joe Perches <[EMAIL PROTECTED]>
> > > Date: Mon, 27 Aug 2007 13:57:42 -0700
> > >
> > > > On Mon, 2007-08-27 at 13:41 -0700, David Miller wrote:
> > > > > From: Johannes Berg <[EMAIL PROTECTED]>
> > > > > Date: Mon, 27 Aug 2007 12:54:09 +0200
> > > > > > #define MAC_FMT "%s"
> > > > > > #define MAC_ARG(a) ({char __buf[18]; print_mac(a, __buf); __buf;})
> > > >
> > > > > I don't think this works.
> > > >
> > > > $ cat test_fmt.c
> > > > #include <stdio.h>
> > > > #include <stdlib.h>
> > >
> > > You're just getting lucky in this test case.
> > >
> > > The language does not allow what you are doing, so you're
> > > playing with fire.
> >
> > What exactly to you think it invalid in this code?
> > I think it's fine (except that I'd chose an u8* for the mac
> > address (first arg in print_mac()).
>
> The __buf[] is used out of scope, therefore it's stack space is
> fair game for the compiler to reuse.
>
> When the compiler sees:
>
> printk(FMT, ({ char __buf[x]; print_mac(a, __buf); __buf;}));
>
> It first all of the printk() argument expressions, first FMT and
> then it evaluates the ({ ... }) argument.
>
> Now that the ({ ... }) expression is done, __buf[] is out of
> scope and illegal to reference.
>
> printk() is now called, with a pointer to an out-of-scope buffer.
> This is illegal.
>
> I don't know how else to explain this to you, I can learn how to
> describe the issue in German if this would help :-)
Oh, not needed. I see the bug and indeed, this is a ticking
timebomb.
I don't use the ({}) notation a lot, so I didn't see this here.
--
Greetings Michael.
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